Problem: $\begin{aligned} f(x)&=\sin(-3x^2-3x+7) \\\\ f'(x)&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $\cos(-3x^2-3x+7)(-6x-3)$ (Choice B) B $\sin(-6x-3)$ (Choice C) C $\cos(-3x^2-3x+7)$ (Choice D) D $-\cos(-3x^2-3x+7)(-6x-3)$
Solution: Since $f$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $f(x)=\underbrace{\sin(~\overbrace{-3x^2-3x+7}^{\text{inner}}~)}_{\text{outer}}$ So if $f(x)=w(u(x))$, then: $\begin{aligned} {u(x)}&={-3x^2-3x+7} &&\text{inner function} \\\\ w(x)&=\sin(x)&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={-6x-3} \\\\ {w'(x)}&={\cos(x)} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} f'(x)&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={\cos({-3x^2-3x+7})} \cdot {(-6x-3)} \end{aligned}$